EECE 290 Exam 2 with Solution, Spring 2011

EECE 290 Analog Signal Processing – Quiz 2 April 9, 2011 4 H iL = 0 2 F 10 Ω iR 1. The initial values at t = 0 of iL and iR are: iL = 0 and iR = K A. Determine dt diL at t = 0. Solution: dt L di v L = , where v = -10K. Substituting, 4 10K L v dt diL = − = − 2.5K A/s. 4 H iL = 0 2 F 10 Ω iR – + v Version 1: K = 1, = = − K dt diL 2.5 -2.5 A/s Version 2: K = 2, = = − K dt diL 2.5 -5 A/s Version 3: K = 3, = = − K dt diL 2.5 -7.5 A/s Version 4: K = 4, = = − K dt diL 2.5 -10 A/s Version 5: K = 5, = = − K dt diL 2.5 -12.5 A/s 2. Determine dt diR at t = 0 in Problem 1. 4 H iL = 0 2 F 10 Ω – + v iC iR Solution: v = -10iR, dt dv dt diR 10 = − 1 ; 2 K C i C i dt dv R C = = = . Substituting, 20 K dt diR = − A/s = -50K mA/s. Version 1: K = 1, = = − K dt diR 50 -50 mA/s Version 2: K = 2, = = − K dt diR 50 -100 mA/s Version 3: K = 3, = = − K dt diR 50 -150 mA/s Version 4: K = 4, = = − K dt diR 50 -200 mA/s Version 5: K = 5, = = − K dt diR 50 -250 mA/s p1/7 3. Which of the following statements is true of the voltage vC across C for large values of t in Problem 1? Version 1: vC approaches zero as an underdamped response Version 2: vC approaches a finite value as an underdamped response Version 3: vC approaches zero as an overdamped response Version 4: vC approaches a finite value as an overdamped response Version 5: vC approaches zero as a critically damped response Solution: The circuit is a parallel circuit; 0 025 2 2 10 1 2 1 = . × × = = p p p C R α ; 0 354 8 1 1 0 = . = = LpCp ω . Since αp is less than ω0, the response is oscillatory. Because of the dissipating element, all responses in the circuit will eventually become zero. Hence, vC approaches zero as an underdamped response. 4. A pulse of amplitude A units and 1 s duration, starting at the origin, is convolved with itself. Determine the value of the convolution integral at t = 1.8 s. Solution: Let the amplitude of the pulse be A. When one pulse is folded about the vertical axis and shifted by t = 1.8 s, the graphical construction becomes as shown. The integration area is y(t) = A×A×0.2 = 0.2A2. 1 λ, s 2 t = 1.8 s 0.8 s A Version 1: A = 1, y(t) = 0.2A2 = 0.2×1 = 0.2 Version 2: A = 1.5, y(t) = 0.2A2 = 0.2×2.25 = 0.45 Version 3: A = 2, y(t) = 0.2A2 = 0.2×4 = 0.8 Version 4: A = 2.5, y(t) = 0.2A2 = 0.2×6.25 = 1.25 Version 5: A = 3, y(t) = 0.2A2 = 0.2×9 = 1.8 5. The delayed triangular pulse of unit height is convolved with the step function Au(t – 2). Determine which of the following choices is a valid convolution integral for t ≥ 5 s. 1 2 1 3 f(t) t, s 1 2 1 3 λ, s -2 t Solution: Let the amplitude of the step be A. When the step is folded about the vertical axis and shifted by t = 5 s, the graphical construction becomes as shown for A = 1. The integration area is A times the area of the triangle, which is 1. Thus, y(t) = Au(t – 5), since this value is valid for t ≥ 5. Version 1: A = 1, y(t) = Au(t – 5) = u(t – 5) p2/7 Version 2: A = 2, y(t) = Au(t – 5) = 2u(t – 5) Version 3: A = 3, y(t) = Au(t – 5) = 3u(t – 5) Version 4: A = 4, y(t) = Au(t – 5) = 4u(t – 5) Version 5: A = 5, y(t) = Au(t – 5) = 5u(t – 5) 4% 6. Given that iSRC = Kδ(t) As. Determine v as a function of t for t ≥ 0+. Solution: The current through R is numerically equal to v, the same as the current of the dependent source. This makes iL = 0 and node a at the same voltage as a’. A current impulse Kδ(t) is thus applied to 1 Ω in parallel with 1 F. At t = 0+, the impulse deposits a charge K coulombs, which gives an initial voltage v(0) = K/1 V across the capacitor. It follows that for t ≥ 0+, . Ke t v t − = ) 0.5 H 1 F 1 Ω – + v v A iSRC 0.5 H 1 F 1 Ω – + v v Kδ(t) iL = 0 v v a a’ ( V Version 1: K = 0.5, V Ke t v t − ( ) = e −t = .5 0 Version 2: K = 1, V Ke t v t − ( ) = e−t = Version 3: K = 1.5, V Ke t v t − ( ) = e −t = .5 1 Version 4: K = 2, V Ke t v t − ( ) = e−t = 2 Version 5: K = 2.5, V Ke t v t − ( ) = e−t = .5 2 7. Determine the quantity of charge that flows through the 1 kΩ resistor, and its direction of flow, from when the switch is closed at t = 0, to t → ∞, assuming v = 1 V. Solution: Ceqs = 1 μF; the charge that flows through the 1 kΩ resistor is the charge on this capacitor, which is q = 1×3v μC. As a check, the initial charges on C1 and C2 are, respectively, 2×v and 2×2v μC. The final charges are 2v – 3v = -v μC and 4v – 3v = v μC, so the voltages are equal and opposite at -v/2 and +v/2. 2 μF 2 μF 1 kΩ – v + – 2v + t = 0 C1 C2 Version 1: v = 1, q = 3v = 3 μC Version 2: v = 2, q = 3v = 6 μC Version 3: v = 3, q = 3v = 9 μC p3/7 Version 4: v = 4, q = 3v = 12 μC Version 5: v = 5, q = 3v = 15 μC 8. Determine the energy dissipated in the resistor from t = 0 to t → ∞ in Problem 7. Solution: Method 1: The power dissipated is that stored in Ceqs, where Ceqs = 1 μF, the initial voltage is 3v and the charge on this capacitor is 3v μC. The stored energy is w = qv/2 = 4.5v2 μJ. Method 2: The initial current through the resistor is 3v/1 = 3v mA. As a function of time, the current is mA, where τ = (1 μF)×(1 kΩ) = 1 ms, so that t is in ms. The energy dissipated is . The dimensions of the integrand are (mA)2×kΩ×ms, which is μJ. The value is [ μJ. τ/ 3 t ve − v ∫ ∞ 0 9 dt e − t × × 2 / 2 1 τ 2 / 4.5 2 e v t − − τ ] 2 0 4.5 v = ∞ Version 1: v = 1, w = 4.5v2 = 4.5 μJ Version 2: v = 2, w = 4.5v2 = 18 μJ Version 3: v = 3, w = 4.5v2 = 40.5 μJ Version 4: v = 4, w = 4.5v2 = 72 μJ Version 5: v = 5, w = 4.5v2 = 112.5 μJ 9. The response of a circuit to a unit step is , where t is in s. Determine the response at t = 10 s to an impulse applied at t = 5 s. /t 10 kte− Solution: The response to an impulse at t = 0 is the time derivative of the response to a unit impulse and is: . The response to an impulse at t = 5 s is . The response at t = 10 s is /10 /10 /10) ( t t te k ke − − − [ ( − 5 /) 10] ( 5 /) 10 1 − − − t ke t [ ] = − − − − 5 /) 10 (10 ke (10 5 /) 10 1 0.5 0.5 ke− . Version 1: k = 1, 0.5 = 0.303 0.5 − ke Version 2: k = 2, 0.5 = 0.607 0.5 − ke Version 3: k = 3, 0.5 = 0.910 0.5 − ke Version 4: k = 4, 0.5 = 1.21 0.5 − ke Version 5: k = 5, 0.5 = 1.52 0.5 − ke ISRC 3 Ω 6 Ω 1 F iR t = 0 10. The switch is closed at t = 0 after being open for a long time. Determine iR as a function of time, assuming ISRC = 1 A. p4/7 Solution: After the switch has been open for a long time, VC = 6ISRC. After the switch is closed, the initial value of iR is -2ISRC. The final value of iR is zero. The time constant, with the switch closed, is (3||6)×1 = 2 s. Hence, iR = /2 2 ISRCe t − − ISRC 3 Ω 6 Ω 1 F iR t = 0 + _ VC Version 1: ISRC = 1 A, iR = = A /2 2 ISRCe t − − /2 2 e−t − Version 2: ISRC = 2 A, iR = = A /2 2 ISRCe t − − /2 4 e−t − Version 3: ISRC = 3 A, iR = = A /2 2 ISRCe t − − /2 6 e−t − Version 4: ISRC = 4 A, iR = = A /2 2 ISRCe t − − /2 8 e−t − Version 5: ISRC = 5 A, iR = = A /2 2 ISRCe t − − /2 10 e−t − 11. A current impulse of K mAs is applied at t = 0, with zero initial conditions. 6% (a) Determine the values of vC and iL at t = 0+. 7% (b) Derive the form of the expression for iL as a function of time for t > 0+ in terms of the two arbitrary constants of the differential equation. Assume that these arbitrary constants are unknown but specify the values of all other quantities in the expression for iL. Kδ(t) 1 kΩ – + vC 0.1 H 0.625 μF iL 7% (c) Determine the values of the arbitrary constants assuming, vC = 200 V and iL = 1 A at t = 0+. Solution: (a) The current source impulse in parallel with R Ω can be transformed to a voltage source impulse of (KR)δ(t) Vs in series with R. As argued in the book for a series RLC circuit, the voltage impulse appears across L and produces a jump in iL equal to KR/L = K×10-3×103/(0.1) = 10K A. The finite jump in iL does not change vC, which remains at zero. Alternatively, it can be argued that the current impulse will flow through R and not through L and C. If the current impulse flows through L and C, vC will jump in value by a finite amount, and the voltage across L will change proportionately to δ(1)(t). The voltage across L and C will be inconsistent with the voltage across R, which can be only be proportional to either δ(t) or to a finite quantity. The current impulse through R will cause a voltage impulse KRδ(t) across R. This impulse will appear across L and cause a jump in current of KR/L. (b) For t > 0+, the source is zero, and the circuit reduces to a series RLC circuit. p5/7 4 10 .1 0.625 0 1 1 6 0 = × × = = − LC ω krad/s; 5 0.1 2 1000 2 = × = = L R α krad/s. Since α > ω0, the response is overdamped, 2 4 5 5 2 2 1 = − − s = − + krad/s and 8 4 5 5 2 2 2 = − − − s = − krad/s. It follows that A, where t is in ms. t t L Be Ae i 8 2 − − + = (c) At t = 0+, A + B = 1; t t e B e A q 8000 2000 8000 2000 − − − = − C, where the constant of integration is zero since q → 0 as t → ∞; at t = 0+, − × − = 1.25 103 A C q vC , 5 B ×103 or 5 10 1.25 10 200 3 3 × − × = − B A , or 4A + B = -1. This gives A = -2/3 A and B = 5/3 A. Alternatively, ( t ) t i L Be Ae dt d L v L 8 2 8 2 0.1 − − − − = = H×A/ms, or ( t ) t L Be Ae v 8 2 8 2 100 − − − − = V, or -vL/100 = 2A + 8B at t = 0+. But vL + vC + vR = 0, or -vL = vC + vR. This gives, 2A + 8B = (200 + 1000)/100. Hence, A + 4B = 6, which gives the same values for A and B. Version 1: (a) K = 0.1; iL(0+) = 10K = 1 A, vC(0+) = 0 Version 2: (a) K = 0.2; iL(0+) = 10K = 2 A, vC(0+) = 0 Version 3: (a) K = 0.3; iL(0+) = 10K = 3 A, vC(0+) = 0 Version 4: (a) K = 0.4; iL(0+) = 10K = 4 A, vC(0+) = 0 Version 5: (a) K = 0.5; iL(0+) = 10K = 5 A, vC(0+) = 0 1 2 1 2 3 3 f(t) t, s 1 2 3 g(t) -1 4 1 2 3 t, s 4 12. Given the two functions f(t) and g(t) shown. 15% (a) Sketch y(t) = f(t)∗g(t) as function of t. 5% (b) Verify y(t) from the product of two polynomials. Solution: (a) When f(t) is folded, it becomes as shown. The convolution integral is zero until t = 3, when side aa’ coincides with g(1). At t = 4, aa’ coincides with g(2), and the area under the product of the two functions is 3×2 = 6. At t = 5, aa’ coincides with g(3), and the area under the product of the two functions is 3×2 – 1×2 = 4. At t = 6, aa’ coincides with g(4), and the area -1 -2 -3 1 2 3 -1 -4 1 2 3 t, s 4 a a’ b b’ p6/7 under the product of the two functions is - 1×2 + 2×2 = 2. At t = 7, bb’ coincides with g(3), and the area under the product of the two functions is 2×2 = 4. At t = 8, bb’ coincides with g(4), and the area under the product of the two functions is back to zero. The convolution function y(t) is as shown. 2 6 1 2 3 t, s 4 5 6 7 8 4 (4,6) (5,4) (7,4) (6,2) y(t) (b) To derive y(t) In terms of a polynomial, f(t) and g(t) are assumed to both start at t = 0. f(t) is divided into two successive 1 s intervals, each of amplitude 2 units. Thus, as polynomials in x, f(x) = 2x + 2, and g(x) = 3x2 – x + 2. The product f(x)g(x) = 6x3 + 4x2 + 2x + 4. The non- zero breakpoints are thus at (1,6), (2,4), (3,2), and (4,4). For the given functions, the total shift in time from the origin is 3 s. Because the system is linear and time-invariant, as assumed in the derivation of the convolution integral, the 3 s is added to the time coordinates of the breakpoints from the product of polynomials to give the non-zero breakpoints of y(t) in the figure. 12 V – + t = 0 2 H 3 H 3 Ω 6 Ω 4 Ω iL1 iR1 13. The switch is moved at t = 0 after having been in the first position for a long time, with iL1 initially zero. Determine, as a function of t, for t ≥ 0+: 10% (a) iR1 10% (b) iL1 Solution: After switching, the circuit becomes as shown, where at t = 0+: iL2 = 12/4 = 3 A, iL1 = 0, iR = 3 A, v = 3×2 = 6 V. As t → ∞, v = 0. The time constant is (2||3)/2 = 1.2/2 = 0.6 s. Hence v = / 0.6 V. 6 t e− p7/7 (a) It follows that /0.6 1 6 t R e v i = − − = − A (b) iL1 can be determined from conservation of flux linkage in the loop formed by the two inductors. At t = 0+, going clockwise around the loop, the total flux linkage is 3×2 + 0 = 6 Wb-T. As t → ∞, the currents in the two inductors are equal in magnitude. Going clockwise around the loop, iL2×5 = 6 Wb-T, so that iL1 = -iL2 = -1.2 A. It follows that ) A. 1.2(1 1.2) (0 1.2 /0.6 /0.6 1 t t L e e i − − − = − + + − = – 2 H 3 H 2 Ω iL1 iL2 iR + v Alternatively, [ ] ( /0.6 ) 0 0.6 / 0 /0.6 1 0 1 1.2 1 2 6 3 1 (0) 1 t t t t t L t L e e dt e i vdt L i − − − − = − = − − = − + = − ∫ ∫ A.